Trying to show that $\ln(x!) - \ln(\lfloor\frac{x}{2}\rfloor!) - \ln(\lfloor\frac{x}{3}\rfloor!) - \ln(\lfloor\frac{x}{6}\rfloor!) \ge \psi(x)$

Question

I've been told that the approach below will not work.

I would be interested if someone could help me to understand what will go wrong.

Let:

$$\psi(x) = \sum\limits_{p^k \le x} \ln p$$

So that (see here):

$$\ln(x!) = \sum_{k=1}\psi(\frac{x}{k})$$

So that:

$$\ln(x!) - \ln\left(\left\lfloor\frac{x}{2}\right\rfloor!\right)- \ln\left(\left\lfloor\frac{x}{3}\right\rfloor!\right)- \ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right) = $$

$$\psi(x) + \psi\left(\frac{x}{5}\right) - 2\psi\left(\frac{x}{6}\right) + \ldots + \psi\left(\frac{x}{6w-5}\right) + \psi\left(\frac{x}{6w-1}\right) - 2\psi\left(\frac{x}{6w}\right) + \ldots$$

So, for each $w > 1$, it follows that:

$$\psi\left(\frac{x}{6w-5}\right) + \psi\left(\frac{x}{6w-1}\right) \ge 2\psi\left(\frac{x}{6w}\right)$$

So, if there exists $k$ such that:

$$\sum_{w=2}^{k} \psi\left(\frac{x}{6w-5}\right) + \psi\left(\frac{x}{6w-1}\right) - 2\psi\left(\frac{x}{6w}\right) \ge 2\psi\left(\frac{x}{6}\right) - \psi\left(\frac{x}{5}\right)$$

Then it follows that:

$$\ln(x!) - \ln\left(\left\lfloor\frac{x}{2}\right\rfloor!\right)- \ln\left(\left\lfloor\frac{x}{3}\right\rfloor!\right)- \ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right) - \psi(x) \ge 0$$

What will be the problem with this approach? Am I making a mistake in any of my steps?

Thanks,

-Larry